The balanced chemical reaction describing this decomposition is as follows:
<span>4c3h5n3o9 .............> 6N2 + 12CO2 +10H2O + O2
From the periodic table:
mass of oxygen = 16 grams
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of carbon = 12 grams
Therefore:
mass of </span><span>C3H5N3O9 = 3(12) + 5(1) + 3(14) + 9(16) = 227 grams
mass of O2 = 2(16) = 32 grams
From the balanced chemical equation:
4(227) = 908 grams of </span>C3H5N3O9 produce 32 grams of O2. Therefore, to know the amount of oxygen produced from 4.5*10^2 grams <span>C3H5N3O9, all we need to do is cross multiplication as follows:
amount of oxygen = (4.5*10^2*32) / (908) = 15.859 grams</span>
moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M
Answer:
Bohrium (Niels Bohr)
Curium (Marie and Pierre Curie)
Einsteinium (Albert Einstein
First, we have to calculate the number of moles of H2SO4 in the solution:
V=60 mL = 0.06 L
c=5.85 mol/L
n=V×c=0.06×5.85=0.351 mol
Then we need to find the molar mass of H2SO4:
2×Ar(H) + Ar(S) + 4×Ar(O) =
=2 + 32 + 64 = 98 g/mol
Finally, we need to find the mass of H2SO4:
m=0.351 × 98 = 34.398 g
Answer:
392 g
Explanation:
The given concentration tells us that<em> in 100 g of solution, there would be 15.3 g of 2-ethyltoluene</em>.
With that in mind we can<u> calculate how many grams of solution would contain 60.0 g of 2-ethyltoluene</u>:
- Mass of solution * 15.3 / 100 = 60.0 g 2-ethyltoluene
